Question: Use the ALEKS graphing calculator to solve the system of equations. \[ y = 2x + 1 \] \[ y =…

Use the ALEKS graphing calculator to solve the system of equations.

\[ y = 2x + 1 \]

\[ y = 2x^2 + 4x - 1 \]

Round to the nearest hundredth.

If there is more than one solution, use the “or” button.

If applicable, click on “No solution”.

\[ (x, y) = (\boxed{\phantom{0}}, \boxed{\phantom{0}}) \]

\[ \text{or} \]

\[ \text{No solution} \]

Solution

To solve the system of equations: \[ \begin{cases} y = 2x + 1 & \\ y = 2x^2 + 4x - 1 & \end{cases} \] Set the two expressions for \(y\) equal to each other: \[ 2x + 1 = 2x^2 + 4x - 1 \] Subtract \(2x + 1\) from both sides: \[ 0 = 2x^2 + 4x - 1 - 2x - 1 \] Simplify the equation: \[ 0 = 2x^2 + 2x - 2 \] Divide every term by 2: \[ 0 = x^2 + x - 1 \] Use the quadratic formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\), where \(a = 1\), \(b = 1\), \(c = -1\): Calculate the discriminant: \[ b^2 - 4ac = 1^2 - 4(1)(-1) = 1 + 4 = 5 \] Calculate the roots: \[ x = \frac{{-1 \pm \sqrt{5}}}{2} \] So, the solutions for \(x\) are: \[ x_1 = \frac{{-1 + \sqrt{5}}}{2} \] \[ x_2 = \frac{{-1 - \sqrt{5}}}{2} \] Find the corresponding \(y\)-values by substituting \(x\) back into \(y = 2x + 1\). For \(x_1 = \frac{{-1 + \sqrt{5}}}{2}\): \[ y_1 = 2\left(\frac{{-1 + \sqrt{5}}}{2}\right) + 1 = -1 + \sqrt{5} + 1 = \sqrt{5} \] For \(x_2 = \frac{{-1 - \sqrt{5}}}{2}\): \[ y_2 = 2\left(\frac{{-1 - \sqrt{5}}}{2}\right) + 1 = -1 - \sqrt{5} + 1 = -\sqrt{5} \] Rounded to the nearest hundredth, the solutions are: First solution: \( x_1 \approx 0.62, \quad y_1 \approx 2.24 \) Second solution: \( x_2 \approx -1.62, \quad y_2 \approx -2.24 \) Thus, the solutions (rounded to the nearest hundredth) are: \( (0.62, 2.24) \) or \( (-1.62, -2.24) \)